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2 years ago

Is muddy water a substance?

Physics

2 answers:

sergejj [24]2 years ago

7 0

Answer: No, I wouldn't imagine so.

Explanation:

Kazeer [188]2 years ago

7 0

Muddy water is not a substance

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Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

svetlana [45]

Let's call $m=565~g=0.565~kg$ the mass of the glider and $m_w=7\cdot12~g =84~g=0.084~kg$ the total mass of the seven washers hanging from the string.
The net force on the system is given by the weight of the hanging washers:
$F_{net} = m_w g$
For Newton's second law, this net force is equal to the product between the total mass of the system (which is $m+m_w$ ) and the acceleration a:
$F_{net}=(m+m_w)a$
So, if we equalize the two equations, we get
$m_w g = (m+m_w)a$
and from this we can find the acceleration:
$a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2$

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2 years ago

How do your results from ray tracing compare to your results from using the thin-lens equation?

EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where;

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$

Hence the focal length of the convex lens is 20cm

7 0

2 years ago

Which is an example of how the body maintains homostasis

strojnjashka [21]

Answer:

lemme see

Explanation:

5 0

2 years ago

Read 2 more answers

A rocket is launched at an angle of 39◦ above the horizontal with an initial speed of 90 m/s. It moves for 7 s along its initial

vagabundo [1.1K]

Answer:

Y=1370.23m

Explanation:

The motion have two moments the first one the time the initial velocity is accelerating then when the engines proceeds to move as a projectile

$a=19 \frac{m}{s^{2} } \\voy=vo*sin(\alpha )\\voy=90*sin(39 )\\y_{o}=0m\\y_{f}=y_{o}+v_{oy}*t+\frac{1}{2}*a*t^{2}\\y_{f}=90*sin(39)*7s+\frac{1}{2}*19\frac{m}{s^{2} }*(7)^{2}\\y_{f}=861.97m$